Question

dansyi the 12.37 When benzophenone is illuminated with ultravioitu IQUIQUE it is excited into a singlet state. This singlet changes rapidly into a triplet, which phosphoresces. Triethylamine acts as a quencher for the triplet. In an experiment in methanol as solvent, the phosphorescence intensity Iphos varied with amine concentration as shown below. A time-resolved laser spectroscopy experiment had also shown that the half-life of the fluorescence in the absence of quencher is 29 us. What is the value of ko? [Q]/(mol dm-3) 0.0010 0.0050 0.0100 Iphos/(arbitrary units) 0.41 0.25 0.16 A1

Answer 1

dansyi the 12.37 When benzophenone is illuminated with ultravioitu IQUIQUE it is excited into a singlet state. This singlet changes rapidly into a triplet, which phosphoresces. Triethylamine acts as a quencher for the triplet. In an experiment in methanol as solvent, the phosphorescence intensity Iphos varied with amine concentration as shown below. A time-resolved laser spectroscopy experiment had also shown that the half-life of the fluorescence in the absence of quencher is 29 us. What is the value of ko? [Q]/(mol dm-3) 0.0010 0.0050 0.0100 Iphos/(arbitrary units) 0.41 0.25 0.16 A1

The shortening of the lifetime of the excited state is called quenching.

Because the Phosphorescence intensity and lifetime are both proportional to the Phosphorescence quantum yield (τ =φ_f /k_f), plots of I_f,0 /I_f and τ₀ /τ (where the subscript 0 indicates a measurement in the absence of quencher) against [Q] should be linear .

we have from kinetics,

I_f,0 /I_f= 1 + τ₀k_Q[Q]
This equation tells us that a plot of I_f,0 /I_f against [Q] should be a straight line with slope τ₀k_Q.Such a plot is called a Stern–Volmer plot.

we take I_f,0 = 1 (arbitrary units).

[Q] (mol dm-3)	0.0010	0.0050	0.0100
If/(arbitrary units)	0.41	0.25	0.16
If,0 /If	2.44	4	6.25

Stern-Volmer plot - If (0) /If 424*x + 1.97 R2 = 0.998 If (0)/ If 0.002 0.004 0.006 0.008 0.01 [Q] , mol/L

from slop of the plot , τ₀k_Q = 424 L/mol

τ₀  = 29μs = 29*10^-6 s

k_Q = 424 L/mol / 29*10^-6 s = 1.46*10⁷ L /mol-s

please comment below if you have any doubts.